pH of $10^{-8}(\mathrm{M}) \mathrm{HCl}$ solution is

Hint : For $10^{-8}(\mathrm{M}) \mathrm{HCl}$ solution, the solution being very dilute, $\mathrm{H}^{+}$ion conc. is not significant as compared to $\mathrm{H}^{+}$ion furnished from $\mathrm{H}_2 \mathrm{O}$. thus, $\mathrm{H}_{\text {total }}^{+}=\mathrm{H}^{+}{ }_{\mathrm{HCl}}+\mathrm{H}^{+}{ }_{\mathrm{H}_2 \mathrm{O}}$ $=10^{-8}+10^{-7}$ (approx) $=1.1 \times 10^{-7}$ (approx) $\begin{aligned} & \mathrm{pH}=-\log \left(1.1 \times 10^{-7}\right) \\ & =\quad 7-\log (1.1)=6.96(\text { approx })\end{aligned}$ $\therefore \quad$ Value of pH is greater than 6 , less than 7